Problem: Find the distance between the point ${(3, 8)}$ and the line $\enspace {y = -\dfrac{1}{2}x + 7}\thinspace$. {1} {2} {3} {4} {5} {6} {7} {8} {9} {\llap{-}2} {\llap{-}3} {\llap{-}4} {\llap{-}5} {\llap{-}6} {\llap{-}7} {\llap{-}8} {\llap{-}9} {1} {2} {3} {4} {5} {6} {7} {8} {9} {\llap{-}2} {\llap{-}3} {\llap{-}4} {\llap{-}5} {\llap{-}6} {\llap{-}7} {\llap{-}8} {\llap{-}9}
Explanation: First, find the equation of the perpendicular line that passes through ${(3, 8)}$ The slope of the blue line is ${-\dfrac{1}{2}}$ , and its negative reciprocal is ${2}$ Thus, the equation of our perpendicular line will be of the form $\enspace {y = 2x + b}\thinspace$ We can plug our point, ${(3, 8)}$ , into this equation to solve for ${b}$ , the y-intercept. $8 = {2}(3) + {b}$ $8 = 6 + {b}$ $8 - 6 = {b} = 2$ The equation of the perpendicular line is $\enspace {y = 2x + 2}\thinspace$ We can see from the graph (or by setting the equations equal to one another) that the two lines intersect at the point ${(2, 6)}$ . Thus, the distance we're looking for is the distance between the two red points. The distance formula tells us that the distance between two points is equal to: $\sqrt{( x_{1} - x_{2} )^2 + ( y_{1} - y_{2} )^2}$ Plugging in our points ${(3, 8)}$ and ${(2, 6)}$ gives us: $\sqrt{( {3} - {2} )^2 + ( {8} - {6} )^2}$ $= \sqrt{( 1 )^2 + ( 2 )^2} = \sqrt{5} $ The distance between the point ${(3, 8)}$ and the line $\thinspace {y = -\dfrac{1}{2}x + 7}\enspace$ is $\thinspace\sqrt{5}$.